Question

已知x^2=x+1,求下列代数式的值:x^5-5x+2

Answer 1

x^2=x+1
x^4=(x^2)^2=(x+1)^2=x^2+2x+1=(x+1)+2x+1=3x+2
x^5=x^4*x=(3x+2)x=3x^2+2x=3(x+1)+2x=5x+3
所以x^5-5x+2=5x+3-5x+2=5
x^2=x+1
x^2-1=x
显然x不等于0，因为把x=0代入不成立
所以两边可以同时除以x
x-1/x=1
两边平方
x^2-2*x*1/x+1/x^2=1^2
x^2-2+1/x^2=1
x^2+1/x^2=3

Answer 2

x^5
=(x^2)^2 *x
=(x+1)^2 *x
=x^3 +2x^2+x
=x(x+1)+2(x+1)+x
=x^2+x+2x+2+x
=x+1+x+2x+2+x
=5x+3
所以原式=5x+3-5x+2
=5
关键是不停地把x^2转换成x+1